What is the time complexity of Maximum Difference Score in a Grid?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Maximum Difference Score in a Grid?

This problem uses Array, Dynamic Programming, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Difference Score in a Grid?

Insight 1: Dynamic Programming is crucial for efficiently exploring possible paths and avoiding redundant calculations. Insight 2: We can maintain a running minimum of visited cells to efficiently calculate the maximum score at each cell.

Which companies ask LeetCode 3148. Maximum Difference Score in a Grid in interviews?

This problem is frequently asked at Intuit.

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Maximum Difference Score in a Grid - LeetCode 3148 Solution

Maximum Difference Score in a Grid - Complete Solution Guide

Maximum Difference Score in a Grid is LeetCode problem 3148, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an m x n matrix grid consisting of positive integers. You can move from a cell in the matrix to any other cell that is either to the bottom or to the right (not necessarily adjacent). The score of a move from a cell with the value c1 to a cell with the value c2 is c2 - c1 . You can start at any cell, and you have to make at least one move. Return the maximum total score you can achieve. Example 1: Input: grid = [[9,5,7,3],[8,9,6,1],[6,7,14,3],[2,5,3,1]] Output: 9 Explanation: We st

Detailed Explanation

The problem requires finding the maximum difference score achievable by moving through a grid of positive integers. You can start at any cell and move to any cell either to the right or below (not necessarily adjacent). The score of a move is the difference between the value of the destination cell and the value of the starting cell. The goal is to maximize the sum of these differences over at least one move.

Solution Approach

The provided solution uses a dynamic programming approach with space optimization. It iterates through the grid row by row and maintains a `dp` array of size `n` (number of columns). `dp[j]` stores the minimum value encountered so far to reach column `j`. In each row, it keeps track of the minimum value encountered from the left (`min_left`) to handle moves that are not strictly from above. The algorithm calculates the score at each cell (i, j) as `grid[i][j] - min_before`, where `min_before` is the minimum value encountered so far to reach that cell, considering both the `dp` array and the `min_left` value. The maximum score is updated accordingly.

Step-by-Step Algorithm

Step 1: Initialize a `dp` array of size `n` with positive infinity. This array represents the minimum value encountered to reach each column in the previous rows.
Step 2: Initialize `max_score` to negative infinity to track the maximum score achieved so far.
Step 3: Iterate through each row `i` of the grid.
Step 4: Initialize `min_left` to positive infinity. This variable keeps track of the minimum value encountered in the current row from the left up to the current column.
Step 5: Iterate through each column `j` of the current row `i`.
Step 6: Calculate `min_before` as the minimum of `dp[j]` (the minimum value to reach column `j` in the previous rows) and `min_left` (the minimum value encountered so far from the left in the current row).
Step 7: If `min_before` is not infinity, calculate the `score` as `grid[i][j] - min_before` and update `max_score` with the maximum value between the current `max_score` and the calculated `score`.
Step 8: Update `current_min` as the minimum of `grid[i][j]` and `min_before`. This represents the minimum value required to get to (i,j) considering the prior values encountered
Step 9: Update `dp[j]` with `current_min`. This ensures that `dp[j]` always holds the minimum value encountered so far to reach column `j` up to the current row.
Step 10: Update `min_left` with `current_min` to keep track of the minimum value encountered from the left in the current row.
Step 11: After iterating through all rows and columns, return `max_score` as an integer.

Key Insights

Insight 1: Dynamic Programming is crucial for efficiently exploring possible paths and avoiding redundant calculations.
Insight 2: We can maintain a running minimum of visited cells to efficiently calculate the maximum score at each cell.
Insight 3: The problem allows non-adjacent moves, which means a simple row-by-row or column-by-column DP is not directly applicable. We need to consider the minimum value encountered so far in *any* previous cell.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: Intuit.

Maximum Difference Score in a Grid | LeetCode 3148

Medium

Array Dynamic Programming Matrix

Time: O(m*n)Space: O(n)

Solution Code

Available in:

import java.util.List;

class Solution {
    public int maxScore(List<List<Integer>> grid) {
        int m = grid.size();
        int n = grid.get(0).size();
        
        double[] dp = new double[n];
        for (int i = 0; i < n; i++) {
            dp[i] = Double.POSITIVE_INFINITY;
        }
        
        double max_score = Double.NEGATIVE_INFINITY;

        for (int i = 0; i < m; i++) {
            double min_left = Double.POSITIVE_INFINITY;
            for (int j = 0; j < n; j++) {
                double min_before = Math.min(dp[j], min_left);
                
                if (min_before != Double.POSITIVE_INFINITY) {
                    double score = grid.get(i).get(j) - min_before;
                    max_score = Math.max(max_score, score);
                }
                
                double current_min = Math.min(grid.get(i).get(j), min_before);
                dp[j] = current_min;
                min_left = current_min;
            }
        }

        return (int) max_score;
    }
}

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Problem Info

Description

Problem 3148: Maximum Difference Score in a Grid - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

Intuit

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3148

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize a `dp` array of size `n` with positive infinity. This array represents the minimum value encountered to reach each column in the previous rows.
2
Step 2: Initialize `max_score` to negative infinity to track the maximum score achieved so far.
3
Step 3: Iterate through each row `i` of the grid.
4
Step 4: Initialize `min_left` to positive infinity. This variable keeps track of the minimum value encountered in the current row from the left up to the current column.
5
Step 5: Iterate through each column `j` of the current row `i`.
6
Step 6: Calculate `min_before` as the minimum of `dp[j]` (the minimum value to reach column `j` in the previous rows) and `min_left` (the minimum value encountered so far from the left in the current row).
7
Step 7: If `min_before` is not infinity, calculate the `score` as `grid[i][j] - min_before` and update `max_score` with the maximum value between the current `max_score` and the calculated `score`.
8
Step 8: Update `current_min` as the minimum of `grid[i][j]` and `min_before`. This represents the minimum value required to get to (i,j) considering the prior values encountered
9
Step 9: Update `dp[j]` with `current_min`. This ensures that `dp[j]` always holds the minimum value encountered so far to reach column `j` up to the current row.
10
Step 10: Update `min_left` with `current_min` to keep track of the minimum value encountered from the left in the current row.
11
Step 11: After iterating through all rows and columns, return `max_score` as an integer.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loops that iterate through the entire grid. The space complexity is determined by the size of the `dp` array, which is proportional to the number of columns in the grid. The space is optimized because we only need to store the minimums of the previous row and the current row's leftmost values.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Explore all possible paths in the grid using recursion. For each path, calculate the total score and keep track of the maximum score.

Time: O(2^(m*n)), in the worst case, each move has two possibilities (down or right-ish), since we can move to any cell right or below.Space: O(m*n) in the worst case for the recursion stack.

Trade-offs: This approach is simple to understand but is highly inefficient for larger grids due to its exponential time complexity. It is suitable only for very small grids.

✨

Key Insights

1Insight 1: Dynamic Programming is crucial for efficiently exploring possible paths and avoiding redundant calculations.
2Insight 2: We can maintain a running minimum of visited cells to efficiently calculate the maximum score at each cell.
3Insight 3: The problem allows non-adjacent moves, which means a simple row-by-row or column-by-column DP is not directly applicable. We need to consider the minimum value encountered so far in *any* previous cell.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: The grid contains all the same numbers. The solution correctly calculates the difference, and `max_score` will be 0 if that number is the minimum value reached first. If all numbers are same, but a larger one is first visited, the algo correctly calculates the minimum negative difference. The problem states that we must perform at least one move.
Edge case 2: All numbers in the grid are in descending order. The solution will find the maximum difference which will most likely be a negative value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the minimum value from all possible paths. A simple row-by-row or column-by-column DP will not work because of the non-adjacent movement constraint.

✗Common mistake 2: Not initializing the `dp` array with positive infinity. This ensures that the first cell encountered will always be considered the minimum value so far.

✗Common mistake 3: Forgetting to update the `min_left` variable correctly in each row. This variable is crucial for considering moves from the left in the current row.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing a path through a supply chain network where the 'score' represents the cost difference between stages. Maximizing the score means maximizing cost reduction.

🏦Real-world scenario 2: Analyzing stock price movements where the 'grid' represents the price at different time points. The goal is to find the maximum profit (difference) achievable by buying and selling at different points in time, with the constraint that you can only buy before selling (move forward in time).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Difference Score in a Grid"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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